Solve (5 questions)?

Not sure we really need partial fractions except in one case:

#(1)#

#int xsinxcosxdx = 1/2 int xsin2x dx#

integrate by parts:

#int xsinxcosxdx = -1/4 int xd (cos2x) #

#int xsinxcosxdx = -(xcos2x)/4 + 1/4 int cos2xdx #

#int xsinxcosxdx = -(xcos2x)/4 + 1/8 sin2x +C#

#int xsinxcosxdx = (sin2x-2xcos2x)/8 +C#

#(2)#

#int x^P/(x(x^(2P)+1)) dx = int x^(P-1)/(x^(2P)+1) dx #

Substitute #t=x^P#, #dt = Px^(P-1)#

#int x^P/(x(x^(2P)+1)) dx = 1/P int dt/(t^2+1) = 1/P arctant+C #

#int x^P/(x(x^(2P)+1)) dx = 1/P arctanx^P +C#

#(3)#

#int lnx/(xsqrt(1+(lnx)^2))dx#

Substitute #t = 1+(lnx)^2#, #dt =2lnx/x dx#

#int lnx/(xsqrt(1+(lnx)^2))dx = int dt/(2sqrt(t)) =sqrtt + C#

#int lnx/(xsqrt(1+(lnx)^2))dx = sqrt(1+(lnx)^2)+C#

#(4)#

#int sqrt(1+e^x)dx#

Substitute #t=sqrt(1+e^x)#, #dt= e^x/(2sqrt(1+e^x))dx =(t^2-1)/(2t)dx#

#int sqrt(1+e^x)dx = 2int t^2/(t^2-1)dt#

#int sqrt(1+e^x)dx = 2int (t^2-1+1)/(t^2-1)dt#

#int sqrt(1+e^x)dx = 2int dt + 2int dt/(t^2-1)#

#int sqrt(1+e^x)dx = 2t + 2int dt/((t-1)(t+1))#

#2/((t-1)(t+1)) = A/(t-1)+B/(t+1)#

#2= A(t+1)+B(t-1) = (A+B)t+(A-B)#

#{(A+B = 0),(A-B=2):}#

#{(A=1),(B=-1):}#

#int sqrt(1+e^x)dx = 2t + int dt/(t-1)- int dt/(t+1)#

#int sqrt(1+e^x)dx = 2t + ln abs (t-1)- ln abs (t+1) +C#

#int sqrt(1+e^x)dx = 2sqrt(1+e^x) +ln ((sqrt(1+e^x)-1)/(sqrt(1+e^x)+1))+C#

#(5)#

#int (x+arcsinx)/sqrt(1-x^2)dx#

#int x/sqrt(1-x^2)dx+int arcsinx/sqrt(1-x^2)dx#

#int x/sqrt(1-x^2)dx = - int (d(1-x^2))/(2sqrt(1-x^2)) = -sqrt(1-x^2)+C#

#int arcsinx/sqrt(1-x^2)dx = int arcsinx d(arcsinx) = 1/2arcsin^2x+C#

#int (x+arcsinx)/sqrt(1-x^2)dx= 1/2arcsin^2x - sqrt(1-x^2)+C#

Here is the answer to #1#. I would rewrite and then use integration by parts. Recall that #sin(2x) = 2sinxcosx#.

#I = int x(1/2sin(2x))dx#

#I = int (xsin(2x))/2dx#

#I = 1/2int xsin(2x) dx#

Now let #u= x# and #dv = sin(2x)dx#. We can immediately see that #du = dx#. Integrating #dv# is a little harder. Letting #n = 2x#, we get #dn = 2dx# and #dx= (dn)/2#. Therefore #v = -1/2cosn = -1/2cos(2x)#

Now recall that integration by parts is

#int udv =uv - int vdu#

#int xsin(2x)dx = -1/2xcos(2x) - int -1/2cos(2x)dx#

#int xsin(2x)dx= -1/2xcos(2x) + int 1/2cos(2x)dx#

#int xsin(2x)dx= -1/2xcos(2x) + 1/4sin(2x) + C#

#1/2int xsin(2x)dx= 1/8sin(2x) - 1/4xcos(2x) + C#

Hopefully this helps!